SOLUTION: A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
I know th
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Question 287482: A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
I know the answer is 59/143 = .413
I don't know the steps of how to solve the problem. Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
I know the answer is 59/143 = .413
I don't know the steps of how to solve the problem.
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P(3<= x <=5) = p(x=3) + P(x=4) + P(x=5)
= 6C3/13C3 + 6C4/13C3 + 6C5/13C3
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= [6C3+6C4+6C5]/13C3
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= [20+15+6]/286
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= 41/286
= 0.1434
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Cheers,
Stan H.
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You can put this solution on YOUR website! A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
The other tutor's solution is wrong.
We get the probability of the complement event
and subtract from 1. The complement event is a committee with
no boys or 1 boy or 2 boys.
1 - P(0 boys OR 1 boy OR 2 boys)
"OR" means "ADD"
1 - [P(0 boys)+P(1 boy)+P(2 boys)]
P(0 boys) = P(5 girls) =
P(1 boy) = P(4 girls AND 1 boy)
"AND" means "MULTIPLY", so
P(1 boy) = P(4 girls AND 1 boy) =
P(2 boys) = P(3 girls AND 2 boys) =
So,
1 - [P(0 boys)+P(1 boy)+P(2 boys)]
becomes:
1 - []=
Edwin
(